Output resistor value, time constant in Crack

Reddart67 · 1195

0 Members and 1 Guest are viewing this topic.

Offline Reddart67

  • Newbie
  • *
    • Posts: 3
on: September 25, 2018, 09:42:29 AM
I see reference to changing the resistance of the output resistor that is parallel to the headphones. For instance:(I didn’t want to reply to this old thread)
Quote
Yeah, a lot of the issue was sourcing larger, more expensive caps.  Also, the charging time of that capacitor needs to be very short.  To achieve this with larger output caps, the 2.49k resistors need to decrease proportionately.  If you go to a 300-ish uf cap and roughly a 1k resistor at the output to keep the charging time short and charging voltage low, this resistance isn't such a big deal with low imedance headphones, but suddenly a high impedance set of cans will "appear" to be a lower impedance load, as 600 ohms in parallel with 1k is suddenly a 375 ohm load.  With the stock circuit, a 600 ohm set of headphones will appear as a 485 ohm load.  Decreasing the load impedance will reduce output levels, so we chose these resistors carefully to provide the best performance with the toughest to drive (normally) headphones. 

A couple of questions:
1. Why doesn’t the headphones resistance serve the function of charging the cap, as it is less resistive than the resistor (usually a 2.49k on a Crack)?

2. Is there a way to calculate optimum resistors in this spot depending on the output capacitor value  and headphone impedance? Or can someone explain or link the theory that would be relevant  to this question?



Offline Paul Birkeland

  • Global Moderator
  • Hero Member
  • *****
    • Posts: 19319
Reply #1 on: September 25, 2018, 10:45:25 AM
1. Why doesn’t the headphones resistance serve the function of charging the cap, as it is less resistive than the resistor (usually a 2.49k on a Crack)?
If you power up the Crack with your headphones plugged in, then both the headphone voice coils and the 2.49K resistors act in parallel while the cap charges.  Without headphones, the 2.49K resistors can still do the job. 

2. Is there a way to calculate optimum resistors in this spot depending on the output capacitor value  and headphone impedance? Or can someone explain or link the theory that would be relevant  to this question?
This isn't quite the right approach.  The question would be why go any bigger on the output coupling caps? 

Paul "PB" Birkeland

Bottlehead Grunt & The Repro Man


Offline Reddart67

  • Newbie
  • *
    • Posts: 3
Reply #2 on: September 25, 2018, 05:53:09 PM
Maybe to run lower than optimal impedance cans?

Mainly my question was to try and learn the theory of how the resistance affects the performance.  I get the HPF aspect of the capacitor/resistor, but I don’t get the “charging” that is referenced in the quoted part of my post.

If 100uf/2.49K is the optimum, that’s great....but I’d like to know “why”.



Offline Paul Birkeland

  • Global Moderator
  • Hero Member
  • *****
    • Posts: 19319
Reply #3 on: September 26, 2018, 05:00:12 AM
The output impedance of the Crack is about 100 Ohms.  This limits performance into low impedance headphones.  If we could put a 1,000uF cap on the outputs and have the amp work well into low impedance headphones, we would just do that.

For 100 ohm headphones, 100uf caps will result in the amp being -3dB at 15Hz.  For the more optimum 300 ohm load, the amp is -3dB at 5Hz with the 100uF cap.  The 2.49K resistor is mostly chosen based on how slowly the 6080 cathode warms up and starts conducting, so the cap charges fairly slowly and there isn't a ton of voltage produced at the headphone jack while that cap is charging.  Once headphones are plugged in, the parallel combination of just about any headphone and a 2.49K resistor will end up being dominated by the headphone impedance. 

If you want to drive low impedance headphones with a Crack, you can add a 6V/3A filament transformer, change the 270R/5W resistors to 130R/10W resistors, change the 220uF/250V caps to 470uF/250V caps, change the 100uF/160V coupling caps to 200uF/160V, change the 3K/10W resistors to 1.5K/20W resistors, then drill another hole in the chassis plate and add a second 6080.

You can also put feedback around the amp to accomplish this.

Paul "PB" Birkeland

Bottlehead Grunt & The Repro Man


Offline Doc B.

  • Administrator
  • Hero Member
  • *****
    • Posts: 9540
    • Bottlehead
Reply #4 on: September 26, 2018, 07:40:24 AM
Quote
If you want to drive low impedance headphones with a Crack, you can add a 6V/3A filament transformer, change the 270R/5W resistors to 130R/10W resistors, change the 220uF/250V caps to 470uF/250V caps, change the 100uF/160V coupling caps to 200uF/160V, change the 3K/10W resistors to 1.5K/20W resistors, then drill another hole in the chassis plate and add a second 6080. You can also put feedback around the amp to accomplish this.

I'll just mentioned here that we also have two other headphone amp kits that are already better optimized for use with low impedance headphones in the S.E.X. and the Mainline kits.

Dan "Doc B." Schmalle
President For Life
Bottlehead Corp.


Offline Reddart67

  • Newbie
  • *
    • Posts: 3
Reply #5 on: September 27, 2018, 06:19:45 AM

If you want to drive low impedance headphones with a Crack, you can add a 6V/3A filament transformer, change the 270R/5W resistors to 130R/10W resistors, change the 220uF/250V caps to 470uF/250V caps, change the 100uF/160V coupling caps to 200uF/160V, change the 3K/10W resistors to 1.5K/20W resistors, then drill another hole in the chassis plate and add a second 6080.

Or how about just adding the filament transformer and running a 6528?



Offline Paul Birkeland

  • Global Moderator
  • Hero Member
  • *****
    • Posts: 19319
Reply #6 on: September 27, 2018, 07:07:33 AM
Or how about just adding the filament transformer and running a 6528?
...and a new HV power transformer, and a new circuit design.  The SEX amp will cost less than such an endeavor and sound quite a bit better.

Paul "PB" Birkeland

Bottlehead Grunt & The Repro Man